ĐKXĐ: x≠0,x≠1,x≠4,x>0
\(Q=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)=\left(\dfrac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a-1}\right)}\right)=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{a-1-a+4}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)Ta có Q>0\(\Rightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0\)
Ta có \(3\sqrt{a}>0\)
Suy ra \(\sqrt{a}-2>0\Rightarrow\sqrt{a}>2\Rightarrow a>4\)
